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\(M=\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
\(M=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(M=\frac{2}{7}-\frac{1}{\frac{7}{2}}=\frac{2}{7}-\frac{2}{7}=0\)
\(Q=2002:\left[\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}.\frac{-\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}\right]=2002:\left[\frac{2.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}.\frac{-\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}\right]=2002:\left[\frac{2}{7}.\frac{-7}{2}\right]=2002.\left(-1\right)=-2002\)
\(Q=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}.\frac{\frac{-7}{6}-\frac{-7}{8}+\frac{-7}{10}}{\frac{2}{6}-\frac{2}{8}+\frac{2}{10}}\)
=>\(Q=\frac{2.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}.\frac{-7.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{2.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
=>\(Q=\frac{2}{7}.\frac{-7}{2}=\frac{2.7.\left(-1\right)}{7.2}=-1\)
=>Q=-1
\(2014:\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1\frac{2}{5}-\frac{7}{9}+\frac{7}{11}}\cdot\frac{1\frac{1}{6}+0,875-0,7}{\frac{1}{3}+0,25-\frac{1}{5}}\right)\)
\(=2014:\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}\cdot\frac{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}\right)\)
\(=2014:\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}\cdot\frac{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{2}{6}+\frac{2}{8}-\frac{2}{10}}\right)\)
\(=2014:\left(\frac{2}{7}\cdot\frac{7\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}\right)}{2\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}\right)}\right)\)
\(=2014:\left(\frac{2}{7}\cdot\frac{7}{2}\right)=2014\)
Câu 2 đây:
\(|x^2+|x-1||=x^2+2\)
\(\Rightarrow\orbr{\begin{cases}x^2+\left|x-1\right|=x^2+2\\x^2+\left|x-1\right|=-x^2-2\left(l\right)\end{cases}}\)
\(\Rightarrow\left|x-1\right|=2\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
a) \(M=\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+0,5}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)
\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{2}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\frac{2012}{2013}\)
\(=\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\right):\frac{2012}{2013}\)
\(=\left(\frac{2}{7}-\frac{2}{7}\right):\frac{2012}{2013}\)
\(=0\)
\(a,\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3+1\right)}\)
\(=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.7}=\frac{2.6}{3.7}=\frac{4}{7}\)
a) Biểu thức A có một số thập phân, ta nên đổi số này thành phân số.
\(A=\frac{-3}{8}.16\frac{8}{17}-0,375.7\frac{9}{17}\)
\(A=\frac{-3}{8}.16\frac{8}{17}-\frac{3}{8}.7\frac{9}{17}\\ =\frac{-3}{8}.\left(16\frac{8}{17}+7\frac{9}{17}\right)\\ =\frac{-3}{8}.\left(16+7+\frac{8}{17}+\frac{9}{17}\right)\\ =\frac{-3}{8}.24=-9\)
b) Ta đổi các số thập phân thành phân số
\(B=\frac{0,6-\frac{1}{3}+\frac{3}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
\(B=\frac{\frac{3}{5}-\frac{1}{3}+\frac{3}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\\ =\frac{3.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
Dễ nhận thấy \(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\ne0\) và \(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\ne0\) nên trong các phân số có tử và mẫu cùng chứa các thừa số khác 0 này ta có thể rút gọn được và đi đến kết quả:
\(B=\frac{3}{7}-\frac{2}{7}=\frac{1}{7}\)
\(M=\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right):\frac{2014}{2015}\)
\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\frac{2014}{2015}\)
\(=\left[\frac{2.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\right].\frac{2015}{2014}\)
\(=\left(\frac{2}{7}-\frac{1}{\frac{7}{2}}\right).\frac{2015}{2014}=\left(\frac{2}{7}-\frac{2}{7}\right).\frac{2015}{2014}=0\)
Ta có M = \(\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right):\frac{2014}{2015}\)
= \(\left(\frac{2\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}{7\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-0,25+\frac{1}{5}\right)}\right):\frac{2014}{2015}\)
= \(\left(\frac{2}{7}-\frac{1}{\frac{7}{2}}\right):\frac{2014}{2015}=\left(\frac{2}{7}-\frac{2}{7}\right):\left(\frac{2014}{2015}\right)=0\)